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Let $S:=\{A_1,A_2,\ldots,A_k\}$ for $A_i\subset\{1,\ldots,n\}$. I'm trying to figure out the correct notation to refer to $S_1:=\{A_1,A_2,\ldots,A_k,A_{k+1}\}$ and $S_2:=\{A_1,A_2,\ldots,A_{k-1}\}$. Is $S_1=S\cup{A_{k+1}}$ and $S_2=S\setminus{A_k}$ or rather $S_1=S\cup\{A_{k+1}\}$ and $S_2=S\setminus\{A_k\}$? I'm just unsure because the $A_i$ are subsets and not elements of $\{1,\ldots,n\}$. Also, if I wanted to denote the family $\{\{1,2\},\{1\}, \{2\},\ldots,\{n\}\}$ would it be $\{1,2\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{i\}\right)$ or $\{\{1,2\}\}\cup\left(\bigcup_{1\leq{i}\leq{n}}\{\{i\}\}\right)$ or something else? Thanks.

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2 Answers 2

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Your second assumption where we had $S_1 = S \cup \{A_{k+1}\}$ and $S_2 = S \setminus \{A_k\}$ is correct. Sure, the $A_i$ are sets themselves, but $S_1$ and $S_2$ are sets of sets. If we want to add or remove elements to (or from) the $S_i$, we also need to add or remove a set that contains sets (the $A_i$) as its elements. A similar logic would be true for your second question. You would want $$\{\{1,2\}\} \cup \left(\bigcup_{i=1}^n \{\{i\}\} \right)$$

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Your second choices are correct in all three cases: $$ S_1 = S \cup \{A_{k+1}\} \\ S_2 = S \setminus \{A_k\} \\ \{\{1,2\},\{1\}, \{2\},\dots,\{n\}\} = \{\{1,2\}\}\cup\left(\bigcup_{i=1}^n\{\{i\}\}\right) $$ But I think the versions with $\dots$ are the clearest. Forget about $S,S_1,S_2$, and instead define $$S_k := \{A_1,A_2,\dots,A_k\}$$

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  • $\begingroup$ Thanks. It's just that in my paper, $S$ is already defined and my proof requires a modification to one of the $A_i$ where $i$ may vary and is not necessarily $k$ in all cases. $\endgroup$
    – Ari
    Feb 4, 2021 at 15:21

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